All 4 members are boys: [tex]\dfrac{\binom{10}4\binom{12}0}{\binom{22}4}=\dfrac6{209}[/tex]
All 4 members are girls: [tex]\dfrac{\binom{10}0\binom{12}4}{\binom{22}4}=\dfrac9{133}[/tex]
At least 1 girl: this outcome is complementary to the event that the committee consists of all boys, so this probability is [tex]1-\dfrac6{209}=\dfrac{203}{209}[/tex].
If you're not familiar with the notation,
[tex]\dbinom nk=\dfrac{n!}{k!(n-k)!}[/tex]
